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Olin [163]
3 years ago
11

Find the similarity ratio of a cube with volume 216 ft to a cube with volume 1000ft

Mathematics
1 answer:
Vika [28.1K]3 years ago
3 0
\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
-----------------------------\\\\

\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-----------------------------\\\\
\cfrac{s^3}{s^3}\implies \cfrac{216}{1000}\implies \cfrac{\sqrt[3]{216}}{\sqrt[3]{1000}}\implies \cfrac{s}{s}\impliedby \textit{similarity ratio}

and simplify it away
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Answer:

a

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b

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c

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d

    The decision rule is  

Reject the null hypothesis

e

There is sufficient evidence to support the researchers claim

Step-by-step explanation:

From the question we are told that

 The first sample size is  n_1 = 30

 The sample variance for elementary school is  s^2_1 = 8324

 The second sample size is  n_2 = 30

  The sample variance for the secondary school is  s^2_2 = 2862

   The significance level is  \alpha = 0.05

The null hypothesis is  H_o :  \sigma^2_1 = \sigma^2 _2

The alternative hypothesis is H_a :  \sigma_1 ^2 > \sigma^2_2

Generally from the F statistics table  the critical value of \alpha = 0.05 at first and  second degree of freedom df_1 = n_1 - 1 = 30 - 1 = 29 and  df_2 = n_2 - 1 = 30 - 1 = 29 is  

         F_{critical} = 1.8608

Generally the test statistics is mathematically represented as

       F = \frac{s_1^2 }{s_2^2}

=>   F = \frac{8324 }{2862}

=>   F = 2.9085

Generally from the value obtained we see that  F >  F_{critical } Hence

   The decision rule is  

Reject the null hypothesis

    The conclusion is  

  There is sufficient evidence to support the researchers claim

   

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